Question

The initial velocity of a particle moving along a straight line is 5 m/s and its retardation is 1 ms-2. The distance moved by the particle in fourth second of its motion is​

Answer 1

Answer

• Distance moved by the particle in fourth second of it's motion is 1.5 m

Given

• Initial velocity , u = 5 m/s
• Retardation =  1 m/s²

To Find

• Distance moved by the particle in fourth second of it's motion

Formula

Distance travelled by the particle in nth second is ,

$\bullet \;\;\; \rm S_n=u+\dfrac{a}{2}[2n-1]$

Solution

Retardation and acceleration are both same . But opposite in direction .

⇒ Acceleration , a = - 1 m/s²

Initial velocity , u = 5 m/s

nth second , n = 4 s

Now , Apply formula .

$\implies \rm S_n=u+\dfrac{a}{2}[2n-1]\\\\\implies \rm S_4=5+\dfrac{-1}{2}[2(4)-1]\\\\\implies \rm S_4=5+\dfrac{-1}{2}[8-1]\\\\\implies \rm S_4=5+\dfrac{-1}{2}[7]\\\\\implies \rm S_4=5-\dfrac{7}{2}\\\\\implies \rm S_4=\dfrac{10-7}{2}\\\\\implies \rm S_4=\dfrac{3}{2}\\\\\implies \rm S_4=1.5\ m$